This document summarizes Flammant's solution for determining stress in a half-space due to a line load. It discusses how Flammant used Boussinesq's solution and superposition to calculate the normal stress component σzz at a point A. The line load is treated as a sequence of infinitesimal point loads. Integrating the stress increment dσzz due to each point load from -∞ to ∞ gives the total stress σzz. Flammant's solution shows that the maximum principal stress occurs on circles centered along the load axis, with the circle radius increasing with decreasing stress.

1. FLAMMANT’S PROBLEM
By
Karamcheti Anantha Venkateswara Prasad
203621011

2.  POINT LOAD
Point load have a finite force acting on a surface of zero area.
 LINE LOAD
Line load acts on an infinitely long line rather than a point.

3.  Consider a line load acting on a surface of a halfspace.

4.  Flammant used Boussinesq solution along with principle of
superposition to solve for stress field in halfspace with line load.
 Consider point A in the figure.
 Since the load is acting on the Y axis from infinity to infinity, we can
have origin at any point on Y axis.
 We will try to find out the σzz component of stress at point A due to
line load.

5.  Consider an elemental length dy of the y axis.

6.  For any point inside the halfspace, including A, the line load ρ
acting on the element dy looks like a point load.
 There will be increment of stress dσzz caused by this point load ρdy.
 dσzz is given by Boussinesq solution.

7.  Now to find stress σzz, we have to integrate both sides
As the line load varies from -∞ to ∞

8.  From the figure
b = (x2+z2)
1/2
y = (b tanØ)
dy = b sec2Ø
We can rewrite as

9.  We can find other components of stress similarly

10.  The line load is effectively a sequence of point loads side by side and we are using
superposition to derive σzz.
 This is possible only when we consider linear elastic theory.
 Flamants solution is one of the applications of Boussinesq solution.
 It is an example of plain strain problem.
 Plain strain problems have 1 spatial direction in which only rigid motions occur.
 As a result certain strain will be identically zero.
 Non zero strain functions are not functions of y.
 A particle, that initially has coordinate y0 in reference configuration will always have
coordinate y0 in any deformed plane; unless and otherwise a rigid translation in the
y direction occurs.

11.  Now let us consider a cylindrical surface of radius b alligned with line load.
 We could carry out an analysis to find tractions that act on this surface by using
above equations.
 After doing analysis we find traction vector
where n is the unit normal vector to the cylindrical surface

12.  What this means is cylindrical surface itself is a principal surface.
 The major principal stress
 The intermediate principal surface is defined by vector n = [0,1,0]
 The intermediate principal stress
 The minor principal surface is perpendicular to the cylindrical surface and to the
intermediate principal surface.
 Minor principal stress = 0

13.  When the vertical component of the traction is integrated along the cylindrical
surface, it equilibriates the applied load ρ.
 Another interesting characteristic of Flammant’s problem is the distribution of the
principal stress in space.
 Consider the locus of points on which major principal stress σ1 is constant.
 It is an equation of a circle with centre on z axis at a depth c beyond the origin and
radius as c.

15.  At each point on this circle σ1 is constant.
 It points directly at origin.
 If we consider larger value for c, circle will be larger as c ∞ (1/ σ1).
 This result gives the pressure bulb in the soil beneath the
foundation.
 It is helpfull in visualizing stress fields.

16. THE END